- #1
kgthatsme
- 2
- 0
Im probably being stupid here but I am confused as to when you can use the impulse formula: I = F*t and when you use F*d = 0.5mv^2 for finding a constant applied force.
If a particle's moving at a constant velocity decelerating to 0m/s in a set distance when acted upon by a constant force f, the force I calculate is double when I use I = F*t (momentum lost = force * time of application) than that of when I use
KE lost = Work done = F*d
e.g m= 3Kg u = 4m/s stopping distance(x)= 5m
Using momentum: I = mv - mu = 12Ns
meanwhile v=4 m/s
therefore dx = 4t by integration
ds = 5m
therefore t = 5/4 s
So using I = F*t
F= I/t
F= 12/1.25
F= 9.6N
Using K.E
KE lost = Work done = F*x
F*x = 0.5mv^2
F*x = 0.5*3*4^2
F*x = 24
x=5m
therefore F= 24/5
F= 4.8 N
Why is this half?
If a particle's moving at a constant velocity decelerating to 0m/s in a set distance when acted upon by a constant force f, the force I calculate is double when I use I = F*t (momentum lost = force * time of application) than that of when I use
KE lost = Work done = F*d
e.g m= 3Kg u = 4m/s stopping distance(x)= 5m
Using momentum: I = mv - mu = 12Ns
meanwhile v=4 m/s
therefore dx = 4t by integration
ds = 5m
therefore t = 5/4 s
So using I = F*t
F= I/t
F= 12/1.25
F= 9.6N
Using K.E
KE lost = Work done = F*x
F*x = 0.5mv^2
F*x = 0.5*3*4^2
F*x = 24
x=5m
therefore F= 24/5
F= 4.8 N
Why is this half?